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The work function of caesium metal is $2.14 \;eV$. When light of frequency $6 \times 10^{14}Hz$ is incident on the metal surface , photoemmision of electron occurs. What is the stopping potential of the emitted photoelectrons ?

$\begin {array} {1 1} (a)\;0.345eV & \quad (b)\;0.345V \\ (c)\;03.45V & \quad (d)\;0.03454V \end {array}$

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For stopping potential $V_{\circ}$, we can write the equation for kinetic energy as :
$K = eV_{\circ}$
So, $V_{\circ}=K/e$
$= (0.345 \times 1.6 \times 10^{-19} ) / ( 1.6 \times 10^{-19} ) = 0.345V$
Ans : (b)

answered Dec 23, 2013
edited Mar 13, 2014

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